RE: Coil test data clarification
Couple more questions:
Can coil inductive reactance be directly measured, or failing that, can it be computed from measurable data? What data points would need to be measured? Would we need to know the output frequency of the alternator to perform the computation?
To the extent that inductive reactance is a kind of resistance, is it a contributor to heat build-up in the coil?
Thanks again,
dd
Later edit: So to close out my understanding, the monotonic decrease in amps with increasing RPM is a function of the increasing inductive reactance within the coil due to increasing RPM?.
Would the "computed resistance" (making up a term) be a value computed by dividing the voltage by the measured amps? For example, for the Black Standard coil V/I for 1500 rpm yields 13.8V/1.60a= 8.63 ohms. At 2000 rpm the computed resistance is 9.72 ohms. Would inductive reactance in ohms be equal to the total computed ohms minus the measured static value?
Thanks again for listening to this ramble...I'm retired now, you see (g).
Couple more questions:
Can coil inductive reactance be directly measured, or failing that, can it be computed from measurable data? What data points would need to be measured? Would we need to know the output frequency of the alternator to perform the computation?
To the extent that inductive reactance is a kind of resistance, is it a contributor to heat build-up in the coil?
Thanks again,
dd
Later edit: So to close out my understanding, the monotonic decrease in amps with increasing RPM is a function of the increasing inductive reactance within the coil due to increasing RPM?.
Would the "computed resistance" (making up a term) be a value computed by dividing the voltage by the measured amps? For example, for the Black Standard coil V/I for 1500 rpm yields 13.8V/1.60a= 8.63 ohms. At 2000 rpm the computed resistance is 9.72 ohms. Would inductive reactance in ohms be equal to the total computed ohms minus the measured static value?
Thanks again for listening to this ramble...I'm retired now, you see (g).
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